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simplify solve for inverse tangent line

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diagonalize (

1	2	1
6	−1	0
−1	−2	−1

)

Solution

P=(

−1	−2	−1
−6	−3	2
13	2	1

), D=(

0	0	0
0	3	0
0	0	−4

), P⁻¹=(

112	0	112
−821	−17	−221
−928	27	328

)

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(

1	2	1
6	−1	0
−1	−2	−1

)

Matrix diagonalization

A matrix A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that A=PDP⁻¹

Find the eigenvalues for (

1	2	1
6	−1	0
−1	−2	−1

): λ=0, λ=3, λ=−4

The diagonal matrix D is composed of the Eigenvalues:

D=(

0	0	0
0	3	0
0	0	−4

)

Find the eigenvectors for (

1	2	1
6	−1	0
−1	−2	−1

): (

−1

−6

), (

−2

−3

), (

−1

)

The Eigenvectors corresponding to the Eigenvalues in D compose the columns of P :

P=(

−1	−2	−1
−6	−3	2
13	2	1

)

Find P⁻¹: (

112	0	112
−821	−17	−221
−928	27	328

)

Verify that A=PDP⁻¹

Solve PDP⁻¹: (

1	2	1
6	−1	0
−1	−2	−1

)

P=(

−1	−2	−1
−6	−3	2
13	2	1

), D=(

0	0	0
0	3	0
0	0	−4

), P⁻¹=(

112	0	112
−821	−17	−221
−928	27	328

)

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1	2	1
6	−1	0
−1	−2	−1

) or choose below

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

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